How small would $|x_0 - a|$ be in order for $f(x)$ to converge to a for Newton's Method I found that $f(x) = \cos(x) + \sin(50x)^2$ has a root $a = \pi/2$. 
Whenever we take our initial value $x_0$ close to a we get convergence, if we are far away from a we do not get convergence to our root. 
However, how do I verify that this is true? How small would $|x_0 - a|$ be in order for $f(x)$ to converge to $a$?  
 A: From one point of view, Newton's method works because a simple root $x_0$ of a function $f$ is an attractive fixed point of the Newton's method function $n(x)=x-f(x)/f'(x)$.  Suppose that $f$ is twice, continuously differentiable and there is an interval $I$ about $x_0$ with the property that $|n'(x)|<r<1$ for all $x\in I$. Then, by the mean value theorem, $n$ will be contractive on $I$. Thus, by the contraction mapping theorem, $n$ will have a unique fixed point in $I$, namely $x_0$. So, the set
$$\{x\in\mathbb R:|n'(x)|<1\}$$
contains a largest interval about $x_0$ with the property that every point in that interval is guaranteed to converge to $x_0$ under iteration.
For this particular problem, $n'(x)$ is
$$n'(x)=\frac{\left(-5000 \sin ^2(50 x)+5000
   \cos ^2(50 x)-\cos (x)\right)
   \left(\sin ^2(50 x)+\cos
   (x)\right)}{(100 \sin (50 x) \cos
   (50 x)-\sin (x))^2}.$$
Here is a plot of $n'(x)$ over the interval
$$\frac{\pi}{2}-\frac{22}{1000}<x<\frac{\pi}{2}+\frac{1}{10000}:$$

Note that $n'(x)<1$ on this interval and there are two spots where $n'(x)=-1$. Those two spots are the endpoints of an interval were we are guaranteed convergence.  To 20 digits of precision, they are
$$
\begin{align}
 a &= 1.5497995938420384751 \: \text{ and} \\
 b &= 1.5708808587249283879.
\end{align}
$$
Note that I am not asserting that we have divergence outside this interval; simply that we are guaranteed convergence within it.
